jika sin²A = 3/5, maka nilai tan 2A = ... [tex]a. \: 2 \sqrt{6} \\ b. \: \ \frac{2}{5} \sqrt{6} \\ c. \: \frac{2}{5 \sqrt{6} } \\ d. \: - \frac{2}{5}
Matematika
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Pertanyaan
jika sin²A = 3/5, maka nilai tan 2A = ...
[tex]a. \: 2 \sqrt{6} \\ b. \: \ \frac{2}{5} \sqrt{6} \\ c. \: \frac{2}{5 \sqrt{6} } \\ d. \: - \frac{2}{5} \sqrt{6} \\ e. \: - 2 \sqrt{6} [/tex]
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[tex]a. \: 2 \sqrt{6} \\ b. \: \ \frac{2}{5} \sqrt{6} \\ c. \: \frac{2}{5 \sqrt{6} } \\ d. \: - \frac{2}{5} \sqrt{6} \\ e. \: - 2 \sqrt{6} [/tex]
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1 Jawaban
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1. Jawaban opangunnie
Sin²A=3/5
Sin A=√3/5
Sin A=√1/25.15
=1/5√15
Sin²A+cos²A=1
3/5+cos²A=1
Cos²A=1-3/5=2/5
Cos A=√2/5
=√1/25.10
=1/5√10
Tan 2A=sin 2A/cos 2A
=2 sin A cos A/cos²A-sin²A
=2.1/5√15.1/5√10 /(2/5-3/5)
=2/25.5√6 /-1/5
=2/5√6. (-5)
=-2√6